Theorem 9.2 (Method 1)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Given : ∆ADB and ∆ABC are triangles on same base AB and between
the same parallels AB and DC
To Prove : 𝑎r (∆ADB) = 𝑎r (∆ABC)
Proof: We know that
Area of triangle = 1/2×𝐵𝑎𝑠𝑒×𝐻𝑒𝑖𝑔ℎ𝑡
Here AB is base
We draw DE ⊥ AB
So, DE is the height of ∆ ABD
∴ Area ∆ ABD = 1/2×𝐵𝑎𝑠𝑒×𝐻𝑒𝑖𝑔ℎ𝑡
= 1/2×𝐴𝐵×𝐷𝐸
Here AB, is base
We draw CF ⊥ AB
So, CF is the height of ∆ ABC
∴ Area ∆ ABC
= 1/2×𝐵𝑎𝑠𝑒×𝐻𝑒𝑖𝑔ℎ𝑡
= 1/2×𝐴𝐵×𝐶𝐹
DE & CF are perpendicular between the same parallel line
∴ DE = CF
So, from (1)
Area ∆ ABD = 1/2×𝐴𝐵×𝐷𝐸
= 1/2×𝐴𝐵×𝐶𝐹
= Area ∆ ABC
∴ Area ∆ ABD = Area of ∆ ABC
Hence proved.
Theorem 9.2
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Given : ∆ADB and ∆ABC are triangles on same base AB and between
the same parallels AB and DC
To Prove : 𝑎r (∆ADB) = 𝑎r (∆ABC)
Construction :
Construct a line through A parallel to be meeting DC at E i.e. AE ∥ BC
& Construct a line through B parallel to AD meeting DC at F i.e. BF ∥ AD
Proof :
From (1)
𝑎r (ABFD) = 𝑎r (ABCE)
2 𝑎r (∆ABD) = 2 𝑎r (∆ABC)
𝑎r (∆ABD) = 𝑎r (∆ABC)
Hence, proved